2016 National Graduate Entrance Examination Mathematics I Past Paper

Approach:

This question tests the determination of the convergence or divergence of an improper integral with parameters. Both ends of the interval of integration are improper points, so the order of the integrand relative to the corresponding power function has to be examined separately at $x\to0^+$ and at $x\to+\infty$.

Steps:

1. At $x\to0^+$, $(1+x)^b\to1$, so the integrand is of the same order as $x^{-a}$, and the convergence of $\int_0^1 x^{-a}dx$ requires $a<1$.
2. At $x\to+\infty$, $x^a(1+x)^b\sim x^{a+b}$, so the integrand is of the same order as $x^{-(a+b)}$, and the convergence of $\int_1^{+\infty}x^{-(a+b)}dx$ requires $a+b>1$.
3. The improper integral converges only if it converges at both ends, so the necessary and sufficient condition is $a<1$ and $a+b>1$, and option (C) is correct.

Answer:

C.

Approach:

This question tests the computation of an antiderivative of a piecewise function and the matching condition at the breakpoint. A candidate function must be differentiable everywhere, so not only must the derivative on each piece be correct, but the function must also be continuous at the breakpoint with both one-sided derivatives existing.

Steps:

1. From $f(x)=\begin{cases}x,&x\le0\\x^2,&x>0\end{cases}$ we know that on the piece $x\le0$ the derivative of the antiderivative should be $x$, and on the piece $x>0$ the derivative should be $x^2$; first rule out the options whose derivatives do not agree.
2. The antiderivative must also be continuous at $x=0$: for option (A) both one-sided limits are $0$; for option (B) the left limit is $0$ while the right limit is $1$; for options (C) and (D) the left limit is $-1$ while the right limit is $0$, so none of them is continuous.
3. Finally check differentiability: for option (A) both the left and the right derivative at $x=0$ are $0$, exactly equal to $f(0)=0$, so only (A) is an antiderivative of $f$.

Answer:

A.

Approach:

This question tests the structural properties of the solutions of a first-order linear differential equation. Two solutions are given, and their difference must be a solution of the corresponding homogeneous equation, so $p(x)$ can first be determined; then one particular solution is substituted back into the equation to determine $q(x)$.

Steps:

1. The difference of the two solutions $2\sqrt{1+x^2}$ is a nonzero solution of the homogeneous equation $y'+p(x)y=0$; from $\sqrt{1+x^2}=e^{-\int p(x)dx}$ we get $\int p(x)dx=-\tfrac12\ln(1+x^2)$, that is $p(x)=-\dfrac{x}{1+x^2}$.
2. Substitute $y=(1+x^2)^2+\sqrt{1+x^2}$ and its derivative $y'=4x(1+x^2)+\dfrac{x}{\sqrt{1+x^2}}$ into the equation $y'+p(x)y=q(x)$.
3. Combining term by term gives $q(x)=4x(1+x^2)+\dfrac{x}{\sqrt{1+x^2}}-x(1+x^2)-\dfrac{x}{\sqrt{1+x^2}}=3x(1+x^2)$, so option (A) is correct.

Answer:

A.

Approach:

This question tests the determination of the extreme points of a function of two variables. The key is to use the limit condition given in the question to expand $f(x,y)$ to its leading term, and then to examine the sign of the values taken by $f$ near the origin along different paths, in order to decide whether the origin is an extreme point.

Steps:

1. From $\lim\limits_{(x,y)\to(0,0)}\dfrac{f(x,y)-xy}{(x^2+y^2)^2}=1$ we get the expansion $f(x,y)=xy+(x^2+y^2)^2+o((x^2+y^2)^2)$.
2. Substituting along the path $y=x$ gives $f(x,x)=x^2+4x^4+o(x^4)>0$ ($x$ sufficiently small and $x\neq0$); substituting along the path $y=-x$ gives $f(x,-x)=-x^2+4x^4+o(x^4)<0$.
3. The values of the function have opposite signs on the two paths, which means that near the origin the function takes both positive and negative values, so $(0,0)$ is not an extreme point and option (A) is correct.

Answer:

A.

Approach:

This question tests the properties of similar matrices. A similarity transformation preserves the transpose, the inverse, and the polynomial combinations built out of them; examine each of the four options in turn to see which conclusion is preserved and which is not, and pick out the wrong one.

Steps:

1. Write $B=P^{-1}AP$ ($P$ invertible); then $B^T=P^TAP^{-T}$, that is $A^T$ and $B^T$ are similar, so option (A) is correct.
2. Similarly $B^{-1}=P^{-1}A^{-1}P$, so $A^{-1}$ and $B^{-1}$ are similar, and option (B) is correct.
3. For option (D): $B+B^{-1}=P^{-1}AP+P^{-1}A^{-1}P=P^{-1}(A+A^{-1})P$, so $A+A^{-1}$ and $B+B^{-1}$ are similar, and option (D) is correct.
4. For option (C): $B+B^T=P^{-1}AP+P^TAP^{-T}$, which in general cannot be written as $P^{-1}(A+A^T)P$, so $A+A^T$ and $B+B^T$ need not be similar, and option (C) is wrong.

Answer:

C.

Approach:

This question tests the use of the standard form of a quadratic form to identify the type of a quadric surface. The key is to use the eigenvalues of the matrix of the quadratic form to determine the number of positive and negative signs in the standard form, and hence to decide whether the surface is an ellipsoid, a hyperboloid, a paraboloid or a cylinder.

Steps:

1. The matrix of the quadratic form $f=x_1^2+x_2^2+x_3^2+4x_1x_2+4x_1x_3+4x_2x_3$ is $A=\begin{pmatrix}1&2&2\\2&1&2\\2&2&1\end{pmatrix}$.
2. The eigenvalues of $A$ are found to be $5,-1,-1$, and after an orthogonal transformation the quadratic form becomes the standard form $5y_1^2-y_2^2-y_3^2$.
3. Setting $f=2$ gives $5y_1^2-y_2^2-y_3^2=2$; the standard form has both positive and negative signs and no zero coefficient, so it represents a hyperboloid, and option (B) is correct.

Answer:

B.

Approach:

This question tests the symmetry and the standardization of the normal distribution. Express the two events with the standard normal distribution function $\Phi$, and then use $\Phi(-t)=1-\Phi(t)$ to bring them to a common form for comparison.

Steps:

1. Standardizing the left-hand side: $P\{X\le\mu-1\}=\Phi\left(-\dfrac{1}{\sigma}\right)$.
2. Standardizing the right-hand side: $P\{Y\ge\mu+1\}=1-\Phi\left(\dfrac{1}{\sqrt{2}\sigma}\right)$.
3. By the symmetry of the normal distribution $\Phi(-t)=1-\Phi(t)$, the equality becomes $\Phi\left(\dfrac{1}{\sigma}\right)=\Phi\left(\dfrac{1}{\sqrt{2}\sigma}\right)$; combined with the condition given in the question this gives $\mu=3$, so option (C) is correct.

Answer:

C.

Approach:

This question tests the computation of the expectation of a statistic of the sample-mean type. $T$ is the arithmetic mean of the $X_i^2$, so it suffices to compute each term with the second-order moment formula $E(X^2)=D(X)+E(X)^2$ and then take the average.

Steps:

1. By the definition of the variance, $E(X_i^2)=D(X_i)+E(X_i)^2=\sigma^2+\mu^2$.
2. By the linearity of expectation, $E(T)=\tfrac1n\sum_{i=1}^nE(X_i^2)=\tfrac1n\cdot n(\sigma^2+\mu^2)=\mu^2+\sigma^2$.
3. Comparing with the four options, only (A) agrees with this conclusion, so the answer is (A).

Answer:

A.

Approach:

This question tests the replacement by equivalent infinitesimals and the asymptotic estimate of an integral. The numerator is an integral with a variable upper limit, so one first has to find the leading term of the integral and then take the quotient with the equivalent expression for the denominator.

Steps:

1. As $x\to0$ we have $t\sin t\sim t^2$, hence $\ln(1+t\sin t)\sim t\sin t\sim t^2$, and the integrand satisfies $t\ln(1+t\sin t)\sim t^3$.
2. Hence $\int_0^x t\ln(1+t\sin t)\,dt\sim\int_0^x t^3\,dt=\tfrac{x^4}{4}$.
3. Also $1-\cos x^2\sim\tfrac{x^4}{2}$, so the original limit is $\dfrac{x^4/4}{x^4/2}=\tfrac12$.

Answer:

$\dfrac{1}{2}$.

Approach:

This question tests the computation of the divergence of a vector field. The divergence is the sum of the partial derivatives of the components with respect to their own variables, so differentiate term by term according to the definition and then add.

Steps:

1. Write the vector field as $A=(x+y+z,\ xy,\ z)$; the three components are $P=x+y+z$, $Q=xy$ and $R=z$.
2. Differentiate term by term: $\dfrac{\partial P}{\partial x}=1$, $\dfrac{\partial Q}{\partial y}=x$, $\dfrac{\partial R}{\partial z}=1$.
3. Adding them gives $\operatorname{div}A=1+x+1=2+x$.

Answer:

$2+x$.

Approach:

This question tests the convergence domain of a power series and the computation of its sum function. First determine the convergence domain by the ratio test, and then use the Taylor series of $e^x$, differentiating term by term and multiplying by $x$, in order to produce the coefficient $n^2$.

Steps:

1. By the ratio test, $\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{(n+1)x}{n^2}\to0<1$ holds for every $x$, so the series converges absolutely for all $x$, the radius of convergence is $R=\infty$, and the convergence domain is $(-\infty,+\infty)$.
2. From $e^x=\sum_{n=0}^\infty\dfrac{x^n}{n!}$, multiplying both sides by $x$ and then differentiating gives $\sum_{n=1}^\infty\dfrac{n x^n}{n!}=xe^x$.
3. Differentiating $xe^x$ once more gives $\sum_{n=1}^\infty\dfrac{n^2 x^{n-1}}{n!}=(1+x)e^x$, and multiplying both sides by $x$ gives the sum function $x(1+x)e^x$.

Answer:

The convergence domain is $(-\infty,+\infty)$, and the sum function is $x(1+x)e^x$.

Approach:

This question tests the use of a Taylor expansion to compute a higher-order derivative. $f'''(0)$ is determined only by the coefficient of the cubic term of $f(x)$ at $0$, so it is enough to expand both parts up to the $x^3$ term.

Steps:

1. Expand $\arctan x=x-\dfrac{x^3}{3}+o(x^3)$.
2. Expand $\dfrac{x}{1+ax^2}=x\left(1-ax^2+o(x^2)\right)=x-ax^3+o(x^3)$.
3. Subtracting gives $f(x)=\left(a-\tfrac13\right)x^3+o(x^3)=\tfrac{2a-1}{3}x^3+o(x^3)$, so $f'''(0)=3!\cdot\tfrac{2a-1}{3}=2(2a-1)=1$, and solving gives $a=\tfrac12$.

Answer:

$\dfrac{1}{2}$.

Approach:

This question tests the computation of an inverse matrix by the adjugate method. First compute the determinant to decide invertibility, then construct the adjugate matrix from the algebraic cofactors, and finally use $A^{-1}=\dfrac{1}{|A|}A^*$ to obtain the result.

Steps:

1. Compute the determinant $|A|=-4\ne0$, so $A$ is invertible and the inverse matrix exists.
2. By the adjugate method, compute the algebraic cofactor of each entry of $A$ one by one and arrange them into the adjugate matrix $A^*$.
3. From $A^{-1}=\dfrac{1}{|A|}A^*=-\tfrac14 A^*$, multiplying every entry of the adjugate matrix by $-\tfrac14$ gives $A^{-1}$.

Answer:

$\begin{pmatrix}\tfrac12&-\tfrac14&-\tfrac14\\[4pt]-\tfrac14&\tfrac12&-\tfrac14\\[4pt]-\tfrac14&-\tfrac14&\tfrac12\end{pmatrix}$.

Approach:

This question tests a binomial distribution derived from a continuous distribution. First compute the probability $p$ of a success in a single trial, and then use the probability formula of the binomial distribution to get the required term.

Steps:

1. From the density $f(x)=2x$ ($0<x<1$) we get $p=P\left\{X\le\tfrac12\right\}=\int_0^{1/2}2x\,dx=\left.x^2\right|_0^{1/2}=\tfrac14$.
2. Since $Y$ denotes the number of occurrences of the event $\{X\le\tfrac12\}$ in 3 independent repeated observations, $Y\sim B(3,p)$, where $p=\tfrac14$.
3. By the probability formula of the binomial distribution, $P\{Y=2\}=C_3^2p^2(1-p)=3\times\tfrac{1}{16}\times\tfrac34=\dfrac{9}{64}$.

Answer:

$\dfrac{9}{64}$.

Approach:

This question tests the conditional probability and the addition formula for probabilities. First use the product formula to obtain $P(AB)$, then solve for $P(B)$, and finally use the addition formula to get the probability of the union.

Steps:

1. By the product formula $P(AB)=P(A)P(B|A)=0.3\times0.4=0.12$.
2. From $P(A|B)=\dfrac{P(AB)}{P(B)}=0.5$ we get $P(B)=\dfrac{0.12}{0.5}=0.24$.
3. Using the addition formula $P(A\cup B)=P(A)+P(B)-P(AB)=0.3+0.24-0.12=0.42$.

Answer:

0.42.

Approach:

This question tests the expectation and the variance of the uniform distribution, together with the numerical characteristics of the sample mean. First write down the expectation and the variance of the population from $X\sim U[0,\theta]$, and then use the expectation and variance formulas for the sample mean to obtain the numerical characteristics of the estimator.

Steps:

1. For $X\sim U[0,\theta]$ we have $E(X)=\tfrac{\theta}{2}$ and $D(X)=\tfrac{\theta^2}{12}$.
2. The sample mean satisfies $E(\overline X)=E(X)=\tfrac{\theta}{2}$ and $D(\overline X)=\tfrac{D(X)}{n}=\tfrac{\theta^2}{12n}$.
3. Hence $E(\hat\theta)=E(2\overline X)=\theta$ and $D(\hat\theta)=4D(\overline X)=\dfrac{\theta^2}{3n}$, that is $\hat\theta$ is an unbiased estimator of $\theta$.

Answer:

$\theta$, $\dfrac{\theta^2}{3n}$.

Approach:

This question tests the total differential and path-independent line integrals. First integrate $\partial f/\partial x$ to obtain $f$ and use the initial condition to fix the function of one unknown, then verify that the field is a gradient field so that the line integral is path independent, and finally evaluate it along a broken-line path and find its extreme value.

Steps:

1. With respect to $x$, integrating $\dfrac{\partial f}{\partial x}$ gives $f(x,y)=xe^{2xy}+C(y)$; substituting the initial condition $f(0,y)=y+1$ gives $C(y)=y+1$, so $f(x,y)=xe^{2xy}+y+1$.
2. Since $\partial f/\partial x\,dy=\partial f/\partial y\,dx$, the vector field is a gradient field and the line integral is path independent, so evaluate it along the broken line from $(0,0)$ to $(1,0)$ and then to $(1,t)$.
3. Adding the contributions along the two segments gives $I(t)=\dfrac12e^{2t}+t$; then finding the extreme value with respect to $t$ gives the minimum value of $I(t)$ as $\dfrac12(1-\ln 2)$.

Answer:

$I(t)=\tfrac{1}{2}e^{2t}+t$, the minimum value is $\tfrac{1}{2}(1-\ln 2)$.

Approach:

This question tests the application of the divergence theorem. A surface integral of the second kind can be converted by the divergence theorem into a triple integral; the integrand is the divergence, which does not depend on the detailed shape of the surface, so only the volume of the region is needed.

Steps:

1. Identify the type of the integral: $I=\iint_{\varSigma}(2x+1)\,dy\,dz+2y\,dz\,dx+3z\,dx\,dy$ is a surface integral of the second kind with $\varSigma$ oriented outwards, and by the divergence theorem it becomes $\iiint_{\varOmega}\operatorname{div}A\,dV$.
2. Compute the divergence: $\operatorname{div}A=\dfrac{\partial(2x+1)}{\partial x}+\dfrac{\partial(2y)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=2+2+3=7$.
3. Compute the volume: $\varOmega$ is enclosed by the plane $2x+y+2z=2$ and the three coordinate planes, so it is a tetrahedron whose intercepts are $1,2,1$, and $V=\tfrac16\cdot1\cdot2\cdot1=\tfrac13$; hence $I=7V=\tfrac73$.

Answer:

$\dfrac{7}{3}$.

Approach:

This question tests the convergence or divergence of a recursively defined sequence. In the first part the difference of two consecutive terms is related to half of the preceding difference, and the comparison with a geometric series shows absolute convergence; in the second part the limit is taken on the basis of that convergence, and the bound on $f'$ excludes both a zero limit and an overly large limit.

Steps:

1. By the mean value theorem, $|x_{n+1}-x_n|=|f(x_n)-f(x_{n-1})|\le\tfrac12|x_n-x_{n-1}|$, hence $|x_{n+1}-x_n|\le\left(\tfrac12\right)^{n-1}|x_2-x_1|$.
2. Comparing with the geometric series $\sum\left(\tfrac12\right)^{n-1}$, the comparison test shows that $\sum_{n=1}^\infty|x_{n+1}-x_n|$ converges absolutely, which proves the first part.
3. By that convergence the sequence $x_n$ converges; let its limit be $a$. Taking limits gives $a=f(a)$; moreover $f(0)=1$ and $f$ is continuous, so $a=0$ does not satisfy $a=f(a)$, and hence $a>0$.
4. From $f(a)-f(0)=a-1$ and the mean value theorem, $a-1=f'(\xi)\cdot a$ ($0<\xi<a$), and since $f'(\xi)<\tfrac12$ we get $a-1<\tfrac a2$, that is $a<2$, and hence $0<a<2$.

Answer:

(I) $\sum_{n=1}^\infty|x_{n+1}-x_n|$ converges absolutely; (II) $\lim_{n\to\infty}x_n$ exists and $0<\lim_{n\to\infty}x_n<2$.

Approach:

This question tests the discussion of the solutions of a system of linear equations with a parameter. First use the coefficient determinant to split the parameter into disjoint cases, and then compare the rank of the coefficient matrix with that of the augmented matrix to decide whether the system has a unique solution, no solution, or infinitely many solutions.

Steps:

1. Compute the coefficient determinant $|A|=(a-1)(-a-2)$, and split the parameter accordingly into the three cases $a=1$, $a=-2$ and $a\neq1,-2$.
2. When $a\neq1,-2$ the matrix $A$ is invertible, $r(A)=r(\bar A)=3$, and the system has a unique solution.
3. When $a=-2$ we have $r(A)=2$ while $r(\bar A)=3$, the two are unequal, and the system has no solution; when $a=1$ we have $r(A)=r(\bar A)=2<3$, and the system has infinitely many solutions.
4. In summary: for $a\neq1,-2$ there is a unique solution; for $a=-2$ there is no solution; for $a=1$ there are infinitely many solutions.

Answer:

$a\neq1,-2$ a unique solution; $a=-2$ no solution; $a=1$ infinitely many solutions.

Approach:

This question tests the diagonalization of a matrix and the computation of its powers, together with the derivation of linear relations among the column vectors from a matrix identity. In the first part the diagonalization is used to turn $A^{99}$ into $\Lambda^{99}$; in the second part $B=BA$ is expanded column by column to obtain the relations among the columns.

Steps:

1. First find the eigenvalues of $A$: from $|A-\lambda E|=0$ the eigenvalues are $0,2,2$, so $A$ is diagonalizable by similarity, $A=P\Lambda P^{-1}$, where $\Lambda=\operatorname{diag}(0,2,2)$.
2. Hence $A^{99}=P\Lambda^{99}P^{-1}=P\operatorname{diag}(0,2^{99},2^{99})P^{-1}$; combining this with $A^2=2A$ gives $A^{99}=2^{98}A$.
3. Expanding $B=BA$ column by column gives the linear relations among the columns, and hence expresses each $\alpha_i$ as a linear combination of the $\beta_i$: $\alpha_1=\beta_1+\beta_3,\ \alpha_2=2\beta_2,\ \alpha_3=\beta_1+\beta_3$.

Answer:

(I) $A^{99}=2^{98}A$;
(II) $\alpha_1=\beta_1+\beta_3,\ \alpha_2=2\beta_2,\ \alpha_3=\beta_1+\beta_3$.

Approach:

This question tests the density of a two-dimensional uniform distribution, the determination of the independence of random variables, and the piecewise computation of a distribution function. The three parts use, in order, the area method to get the density, a counterexample to decide independence by the definition of independence, and a case split according to the range of $z$ and the value taken by $U$ to find the distribution function.

Steps:

1. (I) The area of the region $D=\{(x,y)\mid 0<x<1,\ x^2<y<x\}$ is $\int_0^1(x-x^2)\,dx=\tfrac12$, so the density of the uniform distribution on $D$ is $f(x,y)=\dfrac{1}{|D|}=2$, and it is $0$ elsewhere.
2. (II) To test independence: $U=1$ corresponds to the event $\{XY\le\tfrac14\}$, and the probability given by its area on $D$ is not equal to the product of $P(U=1)$ and $P(X\le x)$, so $U$ and $X$ are not independent.
3. (III) To find $F_Z(z)=P(U+X\le z)$: split according to $z<0$, $0\le z<1$, $1\le z<2$ and $z\ge2$, and within each piece split again according to the two possible values $U=0$ and $U=1$, integrating the distribution of $X$ separately and adding.

Answer:

(I) $f(x,y)=\begin{cases}2,&(x,y)\in D\\0,&\text{other}\end{cases}$;
(II) not independent (the probability of the joint event is not equal to the product);
(III) write $F_Z(z)$ piecewise: it is 0 for $z<0$, for $0\le z<1$ integrate separately over the two cases using $X$ and $U=0/1$, the same holds for $1\le z<2$, and it is 1 for $z\ge2$.

Approach:

This question tests the distribution function of an order statistic and the determination of the coefficient of an unbiased estimator. $T$ is the maximum of the three sample values, so use the general result for order statistics $F_T(t)=F_X(t)^3$; in the second part $E(T)$ is then used to solve for the coefficient $a$.

Steps:

1. First find the distribution function of the population: $F_X(t)=\int_0^t\dfrac{3x^2}{\theta^3}\,dx=\dfrac{t^3}{\theta^3}$ ($0<t<\theta$).
2. (I) The distribution function of $T=\max\{X_1,X_2,X_3\}$ is $F_T(t)=F_X(t)^3=\dfrac{t^9}{\theta^9}$, and differentiating gives the density $f_T(t)=\dfrac{9t^8}{\theta^9}$ ($0<t<\theta$), and $0$ elsewhere.
3. (II) Compute $E(T)=\int_0^\theta t\cdot\dfrac{9t^8}{\theta^9}\,dt=\dfrac{9}{\theta^9}\cdot\dfrac{\theta^{11}}{11}=\dfrac{9}{10}\theta$; setting $E(aT)=aE(T)=\theta$ gives $a=\dfrac{10}{9}$, and then $aT$ is an unbiased estimator of $\theta$.

Answer:

(I) $f_T(t)=\begin{cases}\dfrac{9 t^8}{\theta^9},&0<t<\theta\\0,&\text{other}\end{cases}$;
(II) $a=\dfrac{10}{9}$.